Indicate the equation of the given line in standard form. show all of your work for full credit. the line containing the diagonal, bd, of a square whose vertices are a(-3, 3), b(3, 3), c(3, -3), and d(-3, -3). find two equations, one for each diagonal.

If we draw thm
the diagonals are lines AC and BD

so
slope=changey/changex
for 2 points the slope is (y2-y1)/(x2-x1)
find AC
AC is (-3,3) and (3,-3)
slope=(-3-3)/(3-(-3))=-6/(3+3)=-6/6=-1
slope is -1
and a point is (-3,3)
so
y=mx+b
m=slope
y=-1x+b
sub (-3,3)
3=-1(-3)+b
3=3+b
0=b
y=-1x is the equation for AC

for BD
(3,3) and (-3,-3)
slope is (-3-3)/(-3-3)=1
y=1x+b
find b
(3,3)
3=1(3)+b
3=3+b
0=b
y=x is equation of BD

the diagonals are y=-x and y=x
wait, standard form is ax+by=c
k
y=-x
x+y=0 is thefirst

y=x
x-y=0 is 2nd

the equations are
x+y=0 and x-y=0

X^2 - 4x +(4/2^2) + y^2 + 8y + (8/2^2) = -4 + (4/2^2) + (8/2^2)(x + 2)^2 + (y + 4)^2 = -4 + 4 + 16your center is (-2, -4) and the radius is the sqrt of -4 + 4 + 16 which is 4